A pulley of radius $2$ m is rotated about its axis by a force $F = (20t - 5t^2)$ N (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10\ \text{kg m}^2$, the number of rotations made by the pulley before its direction of motion is reversed, is
Answer: (B) more than $3$ but less than $6$
Concept. A time-varying torque: $\alpha(t)=\dfrac{F(t)R}{I}$. Integrate for $\omega(t)$; the direction reverses when $\omega=0$. Integrate again for the angle turned.
Step 1. Angular acceleration. $\alpha=\dfrac{2(20t-5t^2)}{10}=4t-t^2$.
Step 2. Angular velocity. $\omega=2t^2-\dfrac{t^3}{3}$, which is zero again at $t=6$ s.
Step 3. Angle turned.
$$\theta=\int_0^6\left(2t^2-\frac{t^3}{3}\right)dt=\frac{2(216)}{3}-\frac{1296}{12}=144-108=36\ \text{rad}$$
Step 4. Rotations. $\dfrac{36}{2\pi}\approx5.7$.
Answer. More than 3 but less than 6 (Option B).
Solution by Sreeraj P, M.Sc Physics
